Two bodies are thrown simultaneously from a tower with the same initial velocity $v_0$: one vertically upwards,the other vertically downwards. The distance between the two bodies after time $t$ is

The ratio of times taken by a freely falling body to travel the first $5 \ m$, second $5 \ m$, and third $5 \ m$ distances is:

Galileo's law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, $1: 3: 5: 7 \ldots$]". Prove it.

Water drops fall from a tap onto the floor, $5 \ m$ below, at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground at the instant when the first drop strikes the ground is . . . . . . $m$. $(g = 10 \ m/s^2)$

$A$ ball is dropped from a tower of height $80 \,m$. The time it takes to cover the last $50 \%$ of its fall is (acceleration due to gravity $= 10 \,ms^{-2}$)

$A$ stone is thrown upwards with a velocity $50 \,ms^{-1}$. Another stone is simultaneously thrown downwards from the same location with a velocity $50 \,ms^{-1}$. When the first stone is at the highest point, the velocity of the second stone is $...... \,ms^{-1}$ (Take $g = 10 \,ms^{-2}$).